(a) Explain briefly dielectric strength. (b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light. (i) Identify the wave. (ii) Name one suitable detector for the wave. (iii) Name one source of the wave. (c) An oil drop carrying a charge of 1.0 x 10\(^{-19 }\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\(^{-1}\). Calculate the weight of the oil drop. (d) An RLC series circuit consists of a 100\(\Omega\) resistor, 0.05 H inductor and a 25 \(\mu\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the: (i) impedance; (ii) current. (\(\pi\) = 3.14)

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(a) Explain briefly dielectric strength.
(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.
(i) Identify the wave.
(ii) Name one suitable detector for the wave.
(iii) Name one source of the wave.
(c) An oil drop carrying a charge of 1.0 x 10

^{- 19}

C is found to remain at rest in a uniform electric field of intensity 1200 NC

^{- 1}

. Calculate the weight of the oil drop.
(d) An RLC series circuit consists of a 100

Ω

resistor, 0.05 H inductor and a 25

μ

capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:
(i) impedance;
(ii) current. (

π

= 3.14)

Explanation

a) Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in Vmm

^{- 1}

or Vm

^{- 1}

.
(b)(i) The wave is infra-red wave.
(ii) Suitable detector can be -Infra-red camera;
-Thermopile;
-Bolometer or
-Photographic film.
(iii) Source of the wave can be - Hot bodies or sun

(c) E =

\frac{F}{q} = \frac{W}{q}

W =mF = Eq
= 1200 x 1.0 x 10

^{- 19}

= 1200 x 1.0 x 10

^{- 19}

N

(d) X

_{L}

= 1.2

π f L

= 2 x 3.142 x 50 x 0.50 = 15.7

Ω

_{c}

\frac{1}{2 π f L}

\frac{1}{2 \times 3.14 \times 50 \times 25 \times 10^{- 16}}

= 127.4

Ω

(i) Z =

\sqrt{R^{2} + (X_{1} - X_{c})^{2}}

\sqrt{100^{2} + (127.4 - 15.7)^{2}}

= 149.9

Ω

(ii) I

\frac{V}{Z} = \frac{220}{149.9}

= 1.47A

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